A solid cylinder rolls along the floor. What is the ratio of it's translational kinetic energy to its rotational kinetic energy about an axis parallel to its length and through its center of mass?
To determine the ratio of an object's translational kinetic energy to its rotational kinetic energy, we need to understand the relationship between these two forms of energy for a rolling object.
When a solid cylinder of mass "m," radius "r," and a rotational inertia "I" rolls without slipping along a horizontal surface, the relationship between its translational and rotational kinetic energy can be expressed using the following equation:
\(K_{\text{total}} = K_{\text{translational}} + K_{\text{rotational}}\)
To find the ratio of translational kinetic energy to rotational kinetic energy, we can rewrite this equation as:
\(K_{\text{total}} = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2\)
where:
- \(v\) is the linear velocity of the center of mass of the cylinder.
- \(\omega\) is the angular velocity of the cylinder.
- \(I\) is the rotational inertia of the cylinder, which for a solid cylinder about an axis through its center of mass is given by \(I = \frac{1}{2} m r^2\).
Substituting the expression for \(I\), we have:
\(K_{\text{total}} = \frac{1}{2} m v^2 + \frac{1}{2} \left(\frac{1}{2} m r^2\right) \omega^2\)
Simplifying this expression, we get:
\(K_{\text{total}} = \frac{1}{2} m \left(v^2 + \frac{1}{4} r^2 \omega^2\right)\)
From this equation, we can see that the ratio of the translational kinetic energy to the rotational kinetic energy (\(\frac{K_{\text{translational}}}{K_{\text{rotational}}}\)) can be determined by comparing the \(v^2\) term to the \(\frac{1}{4} r^2 \omega^2\) term.
Therefore, the ratio of the translational kinetic energy to the rotational kinetic energy is:
\(\frac{K_{\text{translational}}}{K_{\text{rotational}}} = \frac{v^2}{\frac{1}{4} r^2 \omega^2} = \frac{4v^2}{r^2 \omega^2}\)
In summary, the ratio of the translational kinetic energy to the rotational kinetic energy for a solid cylinder rolling along a horizontal surface about an axis parallel to its length and through its center of mass is given by \(\frac{4v^2}{r^2 \omega^2}\).
To find the ratio of the translational kinetic energy to the rotational kinetic energy, we need to consider the formulas for each type of kinetic energy.
The translational kinetic energy, KE_trans, is given by the equation:
KE_trans = (1/2) * m * v^2
where m is the mass of the cylinder and v is its linear velocity.
The rotational kinetic energy, KE_rot, is given by the equation:
KE_rot = (1/2) * I * ω^2
where I is the moment of inertia of the cylinder and ω is its angular velocity.
For a solid cylinder, the moment of inertia about an axis parallel to its length and through its center of mass can be calculated as:
I = (1/2) * m * r^2
where r is the radius of the cylinder.
Now, let's calculate the ratio of the translational kinetic energy to the rotational kinetic energy:
Ratio = KE_trans / KE_rot
Substituting the formulas for KE_trans and KE_rot:
Ratio = [(1/2) * m * v^2] / [(1/2) * I * ω^2]
The (1/2) terms cancel out, and substituting the expression for I:
Ratio = (m * v^2) / [(1/2) * (1/2) * m * r^2 * ω^2]
Simplifying:
Ratio = 4 * v^2 / (r^2 * ω^2)
Therefore, the ratio of the translational kinetic energy to the rotational kinetic energy is 4 * v^2 / (r^2 * ω^2).